The number of pairs (x, y) satisfying the equation $$\sin x + \sin y = \sin(x + y)$$ and $$|x| + |y| = 1$$ is

$$\sin x + \sin y = \sin(x + y)$$

or, $$2\sin\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right)=2\sin\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x+y}{2}\right)$$

or, $$2\sin\left(\dfrac{x+y}{2}\right)\left(\cos\left(\dfrac{x-y}{2}\right)-\cos\left(\dfrac{x+y}{2}\right)\right)=0$$

so, either $$\sin\left(\dfrac{x+y}{2}\right)=0$$ ----->(1)

or, $$\cos\left(\dfrac{x-y}{2}\right)-\cos\left(\dfrac{x+y}{2}\right)=0$$ ------>(2)

Using (1), when $$\sin\left(\dfrac{x+y}{2}\right)=0$$, then $$x+y=0$$.

So, $$x=-y$$ ,or, $$\left|x\right|=\left|y\right|$$ 

Using, $$|x| + |y| = 1$$

We can say, $$2\left|x\right|=1$$

or, $$\left|x\right|=\dfrac{1}{2}$$

or, $$x=\pm\ \dfrac{1}{2}$$

So, the ordered pairs will be $$\left(\dfrac{1}{2},-\dfrac{1}{2}\right)$$ and $$\left(-\dfrac{1}{2},+\dfrac{1}{2}\right)$$ ------>(3)

Using (2), $$\cos\left(\dfrac{x-y}{2}\right)-\cos\left(\dfrac{x+y}{2}\right)=0$$

We know, $$\cos\theta\ $$ is an even function.

So, either, $$\dfrac{x+y}{2}=\dfrac{x-y}{2}$$ => $$y=0$$

or, $$\dfrac{x+y}{2}=\dfrac{y-x}{2}$$ =>$$x=0$$

Using, $$|x| + |y| = 1$$

When $$y=0$$, $$\left|x\right|=1$$ => $$x=\pm\ 1$$

And when $$x=0$$, $$\left|y\right|=1$$ => $$y=\pm\ 1$$

So, the ordered pairs are $$\left(0,\pm\ 1\right)$$ and $$\left(\pm\ 1,0\right)$$ ------->(4)

From (3) and (4), we can see there are in total 2+4=6 ordered pairs